introduction to laplace transform analysis

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Introduction to Laplace Transform Analysis

Julius O. Smith III ([email protected])Center for Computer Research in Music and Acoustics (CCRMA)

Department of Music, Stanford UniversityStanford, California 94305

August 11, 2002

Abstract

Contents

1 Introduction to Laplace Transform Analysis 2

2 Existence of the Laplace Transform 2

3 Analytic Continuation 3

4 Relation to the z Transform 5

1



2 EXISTENCE OF THE LAPLACE TRANSFORM

1 Introduction to Laplace Transform Analysis

The one-sided Laplace transform of the function x(t) is defined by

X (s)

∆

= ∞0 x(t)e

−st

dt

where t is real and s = σ + jω is complex. There is also a two-sided Laplace transform

obtained by setting the lower integration limit from 0 to −∞. Since we will be analyzingonly causal 1 linear systems using the Laplace transform, we can use either. However, itis customary in engineering treatments to use the one-sided definition. The one- and two-sided Laplace transforms are also called the unilateral and bilateral Laplace transforms,respectively.

When evaluated along the s = jω axis (i.e., σ = 0), the Laplace transform reduces to theFourier transform :

X ( jω) = ∞0

x(t)e− jωtdt

The Laplace transform can therefore be viewed as a generalization of the Fourier transformfrom the real line (a simple frequency axis) to the complex plane.

One benefit of the more general Laplace transform is the ability to transform signalswhich have no Fourier transform. To see this, we can write the Laplace transform as

X (s) =

∞

0

x(t)e−(σ+ jω)tdt =

∞

0

x(t)e−σt

e− jωtdt.

We can thus interpret the Laplace transform as the Fourier transform of an exponentially enveloped input signal. For σ > 0 (the so-called “right-half plane ” (RHP)), this exponentialweighting forces the Fourier-transformed signal toward zero as t → ∞. As long as the signalx(t) does not increase faster than eBt for some B, its Laplace transform will exist for allσ > B . We make this more precise in the next section.

2 Existence of the Laplace Transform

A function x(t) has a Laplace transform whenever it is of exponential order . That is, there

must be a real number B such that

limt→∞

x(t)e−Bt = 0

As an example, every exponential function Aeαt has a Laplace transform for all finite valuesof A and α. Let’s look at this case more closely.

1A system is said to be causal if its response to an input never occurs before the input is received.

“Introduction to Laplace Transform Analysis,” J.O. Smith



3 ANALYTIC CONTINUATION

The Laplace transform of a causal, growing exponential function

x(t) =

Aeαt, t ≥ 00, t < 0

,

is given by

X (s) =

∞

0

x(t)e−stdt =

∞

0

Aeαte−stdt = A

∞

0

e(α−s)tdt = A

α − se(α−s)t

∞

0

= A

α − se(α−σ− jω)∞ −

A

α − s

=

As−α

, σ > α(indeterminate), σ = α∞, σ < α

Thus, the Laplace transform of an exponential Aeαt is A/(s − α), but this is only defined forre {s} > α.

3 Analytic Continuation

It turns out that the domain of definition of the Laplace transform can be extended by meansof analytic continuation . Analytic continuation is carried out by expanding a function of s ∈ C about all points in its domain of definition, and extending the domain of definition toall points for which the series expansion converges.

In the case of our exponential example

X (s) = Aα − s

, (re {s} > α) (1)

the Taylor series expansion of X (s) about the point s = s0 in the s plane is given by

X (s) = X (s0) + (s − s0)X (s0) + (s − s0)2X (s0)

2 + (s − s0)3

X (s0)

3! + · · ·

∆=

∞n=0

(s − s0)nX (n)(s0)

n!

where, writing X (s) as (α − s)−1 and using the chain rule for differentiation,

X (s0) ∆

= X (1)(s0) ∆=

dX (s)ds

s=s0

= (−1)(α − s)−2(−1)s=s0

= 1

(α − s)2

X (s0) ∆

= X (2)(s0) ∆=

d2X (s)

ds2

s=s0

= (−2)(α − s)−3(−1)s=s0

= 2

(α − s)3

X (s0) ∆

= X (3)(s0) ∆=

d3X (s)

ds3

s=s0

= (−3)(2)(α − s)−4(−1)s=s0

= 3!

(α − s)4




3 ANALYTIC CONTINUATION

and so on. We also used the factorial notation n! ∆= n(n − 1)(n − 2) · · · 3 · 2 · 1, and we defined

the special cases 0! ∆= 1 and X (0)(s0)

∆= X (s0), as is normally done. The series expansion of

X (s) can thus be written

X (s) = 1

α − s0+

s − s0

(α − s0)2 +

(s − s0)2

(α − s0)3 + · · ·

=∞n=0

(s − s0)n

(α − s0)n+1 (2)

We now ask for what values of s does the series Eq. (2) converge . The value s = α isparticularly easy to check, since

X (α) =∞n=0

(α − s0)n

(α − s0)n+1 =

∞n=0

1

α − s0= ∞

1

α − s0

Thus, the series clearly does not converge for s = α, no matter what our choice of s0 mightbe. We must therefore accept the point at infinity for H (α). This is eminently reasonablesince the closed form Laplace transform we derived, H (s) = 1/(α − s) does “blow up” ats = α. The point s = α is called a pole of H (s) = 1/(α − s).

More generally, let’s apply the ratio test for the convergence of a geometric series. Sincethe nth term of the series is

(s − s0)n

(α − s0)n+1

the ratio test demands that the ratio of term n +1 over term n have absolute value less than1. That is, we require

1 >

(s − s0)n+1

(α − s0)n+2

(s − s0)n

(α − s0)n+1

=

s − s0α − s0

or,

|s − s0| < |α − s0|

We see that the region of convergence is a circle about the point s = s0 having radiusapproaching but not equal to |α − s0|. Thus, the circular disk of convergence is centered ats = s0 and extends to, but does not touch, the pole at s = α.

The analytic continuation of the domain of Eq. (1) is now defined as the union of the

disks of convergence for all points s0 = α. It is easy to see that a sequence of such disks canbe chosen so as to define all points in the s plane except at the pole s = α.

In summary, the Laplace transform of an exponential x(t) = Aeαt is

X (s) = A

s − α

and the value is well defined and finite for all s = α.




REFERENCES

Analytic continuation works for any finite number of poles of finite order,2 and for aninfinite number of distinct poles of finite order. It breaks down only in pathological situationssuch as when the Laplace transform is singular everywhere on some closed contour in thecomplex plane. Such pathologies do not arise in practice, so we need not be concerned aboutthem.

4 Relation to the z Transform

The Laplace transform is used to analyze continuous-time systems. Its discrete-time coun-terpart is the z transform:

X d(z ) ∆=

∞n=0

xd(nT )z −n

If we define z = esT , the z transform becomes proportional to the Laplace transform of a

sampled continuous-time signal:

X d(esT ) =∞n=0

xd(nT )e−snT

As the sampling interval T goes to zero, we have

limT →0

X d(esT )T = lim∆t→0

∞n=0

xd(tn)e−stn∆t =

∞

0

xd(t)e−stdt = X (s)

where tn∆

= nT and ∆t ∆

= tn+1 − tn = T .

References

[1] P. Denbigh, System Analysis and Signal Processing , Addison-Wesley, 1998, A fair amountof MATLAB code... isn’t cheap: $81.95 (as of 4/01), but it might be a good book forthose rusty with Laplace transform analysis.

[2] Boyce and Deprima, Elementary Differential Equations and Boundary Value Problems ,

John Wiley and Sons, Inc., 1986, Includes a nice mathematical discussion of the Laplacetransform.

2The order of a pole is its multiplicity. For example, the function

H (s) = 1

(s − p)3

has a pole at s = p of order 3.




REFERENCES

[3] W. R. LePage, Complex Variables and the Laplace Transform for Engineers , Dover, NewYork, 1961.

[4] D. K. Frederick and A. B. Carlson, Linear Systems in Communication and Control , JohnWiley and Sons, Inc., 1971.


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